A and B throw a dice alternately til any one gets a 6 on the and win the game. If A begins the game find the probabilities of winning the game.

Explanation

To solve the problem of finding the probabilities of A and B winning the game when they throw a dice alternately until one of them gets a 6, we can break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Probabilities**:
- The probability of rolling a 6 on a die is \( P(6) = \frac{1}{6} \).
- The probability of not rolling a 6 is \( P(\text{not } 6) = \frac{5}{6} \).

2. **Calculating A's Winning Probability**:
- A starts the game. If A rolls a 6 on the first turn, A wins immediately. The probability of this event is:
\[ P(A \text{ wins on first turn}) = \frac{1}{6} \]

3. **Considering the Second Turn**:
- If A does not roll a 6 (which happens with probability \( \frac{5}{6} \)), then it is B's turn.
- If B rolls a 6, B wins. The probability of this event is:
\[ P(B \text{ wins on first turn}) = P(A \text{ does not win}) \times P(B \text{ wins}) = \frac{5}{6} \times \frac{1}{6} \]

4. **Continuing the Game**:
- If both A and B do not roll a 6 in their first turns, the game resets to the original state, but now A has already rolled once. The probability of this scenario is:
\[ P(A \text{ does not win}) \times P(B \text{ does not win}) = \frac{5}{6} \times \frac{5}{6} = \left(\frac{5}{6}\right)^2 \]
- In this case, it is again A's turn, and the game continues. The probability of A winning from this point is still \( P(A) \).

5. **Setting Up the Equation**:
- We can express the total probability of A winning as:
\[ P(A) = P(A \text{ wins on first turn}) + P(A \text{ does not win}) \times P(B \text{ does not win}) \times P(A) \]
- Substituting the known probabilities:
\[ P(A) = \frac{1}{6} + \left(\frac{5}{6}\right)^2 P(A) \]

6. **Solving for \( P(A) \)**:
- Rearranging the equation:
\[ P(A) - \left(\frac{5}{6}\right)^2 P(A) = \frac{1}{6} \]
- Factor out \( P(A) \):
\[ P(A) \left(1 - \left(\frac{5}{6}\right)^2\right) = \frac{1}{6} \]
- Calculate \( 1 - \left(\frac{5}{6}\right)^2 \):
\[ 1 - \frac{25}{36} = \frac{36 - 25}{36} = \frac{11}{36} \]
- Thus, we have:
\[ P(A) \cdot \frac{11}{36} = \frac{1}{6} \]
- Solving for \( P(A) \):
\[ P(A) = \frac{1}{6} \cdot \frac{36}{11} = \frac{6}{11} \]

7. **Finding B's Winning Probability**:
- Since A and B are the only players, the probability of B winning is:
\[ P(B) = 1 - P(A) = 1 - \frac{6}{11} = \frac{5}{11} \]

### Final Probabilities:
- The probability that A wins is \( \frac{6}{11} \).
- The probability that B wins is \( \frac{5}{11} \).